Factor completely. $486 +108 x + 6 x^2=$
Solution: First, we take a common factor of $6$. $486 +108 x + 6 x^2=6(81 +18 x +x^2)$ Now, let's factor $81 +18 x +x^2$. Both $81$ and $x^2$ are perfect squares, since $81=({9})^2$ and $x^2=({x})^2$. Additionally, $18 x$ is twice the product of the roots of $81$ and $x^2$, since $18 x=2({9})({x})$. $81 +18 x +x^2 = ({9})^2 + 2({9})({x})+({x})^2$ So we can use the square of a sum pattern to factor: ${a}^2 + 2( a)( b)+ {b}^2 =({a} + {b})^2$ In this case, ${a}={9}$ and ${b}={x}$ : $ ({9})^2 + 2({9})({x})+({x})^2 =({9} +{x})^2$ $\begin{aligned} 486 +108 x + 6 x^2 &=6(81 +18 x +x^2) \\\\ &=6(9 + x)^2 \end{aligned}$ In conclusion, the complete factorization is $6(9 + x)^2$ Remember that you can always check your factorization by expanding it.